Groups with $G^n \cong G$ for some integer $n$ [duplicate]

Question

Which integers $n>2$ have the following property?

There is a group $G$ such that

• $G^n \cong G$; and
• for all integers $k$ with $1<k<n$ we have $G^k\not \cong G$.

Answer 1

This is possible with abelian groups for any $n$; see this answer to a very similar question.